Proving an identity involving binomial coefficients and fractions I've been trying to prove the following formula (for $n > 1$ natural, $a, b$ non-zero reals), but I don't know where to start.
$$\sum_{j=1}^n \binom{n-1}{j-1} \left( \frac{a-j+1}{b-n+1} \right) \left( \frac{a}{b} \right)^{j-1} \left( \frac{b-a}{b} \right)^{n-j} = \frac{a}{b}$$
Wolfram says it's right, but so far I've been unable to give a proof. I wonder if there's a combinatorial proof to it. Any help, hint or reference will be much appreciated.
 A: Replace $j$ by $j+1$ and $n$ by $n+1$ rewrite as Greg Martin suggested:
\begin{align*}
  \sum_{j=0}^n \binom nj \left( \frac{a-j}{b-n} \right) \left( \frac{a}{b} \right)^j \left( \frac{b-a}{b} \right)^{n-j} \tag I
\end{align*}
Let $p=\dfrac{a}{b}$ and $q=1-p$ and consider:
\begin{align*}
  \left(q+px\right)^n &= \sum_{j=0}^{n}\binom nj p^j q^{n-j} x^j\tag 1 \\
  npx\, \left(q+px\right)^{n-1} &= \sum_{j=0}^{n}j \binom nj p^j q^{n-j} x^j\tag 2 \\
\end{align*}
$(1)$ is by binomial expansion, $(2)$ is by differentiating w.r.t x on both sides.
Now substituting $x=1$, $(1)$ and $(2)$ become
\begin{align*}
  1 &= \sum_{j=0}^{n}\binom nj p^j q^{n-j}\tag 3 \\
  np &= \sum_{j=0}^{n}j \binom nj p^j q^{n-j} \tag 4 \\
\end{align*}
From $(\mathrm{I})$,$(3)$ and $(4)$,
\begin{align*}
  \sum_{j=0}^n \binom nj \left( \frac{a-j}{b-n} \right) \left( \frac{a}{b} \right)^j \left( \frac{b-a}{b} \right)^{n-j} &= \frac{a}{b-n}\cdot 1 - \frac{1}{b-n}\cdot np \\
  &= \frac{a\left(b-n\right)}{b-n} \\
  &= \frac{a}{b}
\end{align*}
